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As usual, I have a ridiculously specific question about a tiny detail of a tiny calculation. There is a tremendous amount of context, so fair warning. I don't need a solution to this whole problem. I need a solution to one specific part, and it is a minor detail. I'll show what I've got so far, and point out my concern at the end.

Our problem is to use residue calculus to compute: $$\int\limits_{-\infty}^{\infty}\frac{\cos(2x)-1}{x^2}dx.$$

We note that this is equal to the real part of the easier-to-work-with integral: $$\int\limits_{-\infty}^{\infty}\frac{e^{2ix}-1}{x^2}dx,$$ and we use a "hippo's back" contour, formed by the positively-oriented half-circle $|z|\leq R$ in the upper half-plane and the line along the real axis from $-R$ to $R$, interrupted from $-\varepsilon$ to $\varepsilon$ by the negatively oriented half-circle in the upper half-plane of radius $\varepsilon$, so as to 'jump' over the simple pole at $z=0$.

Since this contour contains no poles or singularities, etc., the integral of our integrand around it is zero by Cauchy's Theorem. It is also a simple matter to show that the large half-circle's contribution is zero in the limit $R\to\infty$. So now we are left with the fact that in the limit $\varepsilon\to 0$, we obtain:

$$\int\limits_{-\infty}^{\infty}\frac{e^{2ix}-1}{x^2}dx=\lim\limits_{\varepsilon\to 0}\int\limits_{C_\varepsilon}\frac{e^{2iz}-1}{z^2}dz,$$

where $C_\varepsilon$ is the little negatively-oriented half-circle mentioned above.

Now, I know the trick in which the value of the right-hand side happens to give exactly $\pi i \text{Res}(f,0)$ (that is, half of what the full circle would give), and can use it to show that the result is $-2\pi$.

Okay, thanks for reading this far. Here is my actual question: Without resorting to tricks like the half-residue thing, how can we get this result from the right-hand side? That is, how do we prove rigorously that:

$$\lim\limits_{\varepsilon\to 0}\int\limits_{C_\varepsilon}\frac{e^{2iz}-1}{z^2}dz=-2\pi$$

I have been using the parametrization $z=\varepsilon e^{-it}$ with $t\in[0,\pi]$, where the negative in the exponential deals with the negative orientation, which almost always works in these situations, but not here, since the $\varepsilon$ does not entirely cancel from the denominator then as it usually does in these types of problems.

I have confirmed the result with Wolfram-Alpha using the original problem statement, and also noted a few other things, like that the original integrand can be expressed as $\frac{\sin^2(x)}{x^2}$, but gotten nowhere with that.

In a pinch, I'll use the 'half-residue' trick, but being able to do this simply from basic principles would be nice.

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The parametrization works (I'm not sure what you mean about it not fully cancelling out of the denominator). For the negatively oriented half circle, you have $$-\lim_{\epsilon\to 0} \int_0^{\pi}\frac{1}{\epsilon}ie^{-it}(1-e^{2i\epsilon e^{it}})dt$$ and you can expand $(1-e^{2i\epsilon e^{it}}) = -2i\epsilon e^{it}+O(\epsilon^2)$ and get $$-\lim_{\epsilon\to 0}\int_0^\pi (2+O(\epsilon)) dt = -2\pi.$$

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  • $\begingroup$ Thanks, I'm on my way to bed but I will look at this closely in the morning. $\endgroup$ – The Count Jan 29 '17 at 3:05
  • $\begingroup$ The expansion is what does it. I've never thought that way about these types of problems before. FYI, I used a negative exponent in my parametrization instead of the negative out front, but this solves my quandry. Thanks again. Accepted. $\endgroup$ – The Count Jan 29 '17 at 19:58
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For a given, $a$ we have,

$$\int_{-\infty}^\infty \frac{1-\cos(ax)}{x^2}dx =2\int_0^\infty \frac{1-\cos(ax)}{x^2}dx =-2\int_0^\infty \left(\frac{1}{x}\right)'(1-\cos(ax))dx \\=-2\left[ \frac{1-\cos(ax)}{x}\right]^\infty_0 +2a\int_0^\infty\frac{\sin(ax)}{x}dx = 2a\int_0^\infty\frac{\sin(ax)}{x}dx =\color{red}{\pi |a|}$$

Provided that:

$$\lim_{x\to 0} \frac{1-\cos(ax)}{x} = \lim_{x\to 0} a^2x\frac{1-\cos(ax)}{(ax)^2} =0*\frac12 =0.$$

and $$ \int_0^\infty\frac{\sin(ax)}{x}dx =\overset{u=ax}{=}sign(a)a\int_0^\infty\frac{\sin(x)}{x}dx $$

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